I am interested in bounds on the minimal distance between the spectral abscissa $\max_{\lambda\in\sigma(A)}\mathrm{Re}\lambda$ of a matrix $A$ and the eigenvalues of its perturbated version $A+S$. In my case, the matrix $A$ is symmetric and the perturbation $S$ is skew-symmetric. The matrix has eigenvalues with algebraic multiplicity higher than 1 (this is an unnice property that excludes some results). The matrix is a $N$-block matrix with $n\times n$ blocks and the perturbation is block diagonal with $n\times n$ blocks.

A known result in the literature states:

*If $A$ and $B$ are normal with $\|A-B\|=\varepsilon$ in a unitarily invariant norm $\|\cdot\|$ (e.g. the Frobenius norm), then for any eigenvalue $\lambda$ of $A$ there is an eigenvalue $\mu$ of $B$ such that $|\lambda-\mu|\leq\varepsilon$.*


Let $A$ be symmetric and $\nu=\arg\max_{\lambda\in\sigma(A)}\mathrm{Re}\lambda$ (eigenvalue with spectral abscissa of $A$). Let $B=A+S$ where $S$ is skew-symmetric. By the above result, there exists an eigenvalue $\mu$ of $A+S$ such that

$$
|\mathrm{Re}\nu-\mathrm{Re}\mu|\leq|\nu-\mu|\leq\|S\|.
$$ 

I wonder if there is some results that utilises that one matrix is symmetric and the other skew-symmetric and not just normality? 

I've found many results about symmetric (Hermitian) matrices, but not the symmetric and skew-symmetric combination. My intuition says that the skew-symmetric matrix should contribute to the imaginary part of the eigenvalue at first. Therefore, the minimum distance between the spectral abscissa of $A$ and the spectrum of $A+S$ is reduced compared to the case when only normality holds. The idea is that as the spectrum of $A+S$ varies continuously with $S$, it first leaves the real line and goes into the complex plane. Does this seem resonable?